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LXXII.4 (1995)

Class numbers of certain real abelian fields

by

Jae Moon Kim (Inchon)

0. Introduction. We fix an odd prime q. Let p be an odd prime such that p ≡ 1 mod q. Let

g(t) = (1 − t)

p−2

+ 1

2 (1 − t)

p−3

+ . . . + 1 p − 1 .

We will consider g(t) as an element in F

p

[t], where F

p

is the finite field with p elements. If necessary, we will also view g(t) as a polynomial in Z

p

[t], where Z

p

is the ring of p-adic integers. It is not hard to see that

tg(1 − t) ≡ (1 − t)

p

− (1 − t

p

)

p mod p.

This polynomial

f (t) = (1 − t)

p

− (1 − t

p

) p

in F

p

[t] was first introduced by D. Mirimanoff around 1905 and has been exhaustively studied since then. For instance, he used the polynomial f (t) to prove the following striking criterion of A. Wieferich: if the Fermat quotient (2

p−1

− 1)/p is not congruent to 0 mod p, then the first case of the Fermat’s Last Theorem is true (see [8]).

In this paper, we will study class numbers of certain real abelian fields by using the polynomial g(t). Our work is based on the observation that g(t) comes from a Coates–Wiles series. To be precise, let

h

t

(x) = Y

w∈R

((1 + x)

w

− t),

where R = {w ∈ Z

p

| w

p−1

= 1} is the group of the (p − 1)th roots of 1 in Z

p

. Viewing h

t

(x) as an element of Z

p

[t][[x]], i.e., as a power series in x

This work was partially supported by the Basic Science Research Institute program, Ministry of Education,

#

BSRI-94-1414.

[335]

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with coefficients in Z

p

[t], we have the following expansion (see [6]):

h

t

(x) = (1 − t)

p−1

+ g

1

(t)x

p−1

+ (higher terms) with

g

1

(t) ≡ g(t) mod p.

To see why h

t

(x) is a Coates–Wiles series, let t = s be a root of 1 in Z

p

. Then h

s

(x) is indeed a Coates–Wiles series (see [1], [10]).

In Section 1 of this paper we will use the above expansion of h

t

(x) to fac- torize certain principal ideals into a product of prime ideals. In Section 2, we discuss class numbers of certain real abelian fields. Before we state the main theorems, we first explain several notations that will be used throughout this paper.

For each integer n ≥ 1, we choose a primitive nth root ζ

n

of 1 so that ζ

mm/n

= ζ

n

whenever n | m. Let k

0

= Q(ζ

p

), k

n

= Q(ζ

pn+1

), K

0

= Q(ζ

pq

) and K

n

= Q(ζ

pn+1q

). We denote the unique subfield of k

n

of degree p

n

over Q by Q

n

. Let F

0

= Q(ζ

q

), F

n

= Q

n

q

), F

0+

= Q(ζ

q

+ ζ

q−1

) and F

n+

= Q

n

q

+ ζ

q−1

). Thus for E = k, K, F and F

+

, E

n

is the nth layer of the basic Z

p

-extension of E

0

. We denote the Galois groups Gal(F

0

/Q) and Gal(F

0+

/Q) by ∆ and ∆

+

, respectively, and use the same letters ∆, ∆

+

for those Galois groups isomorphic to Gal(F

0

/Q), Gal(F

0+

/Q). Elements of ∆

+

and ∆ will be arranged as ∆

+

= {τ

1

, τ

2

, . . . , τ

l

= id} and ∆ = {±τ

1

, ±τ

2

, . . . , ±τ

l

} with l =

12

ϕ(q). For each τ ∈ ∆, let p(τ ) be the integer modulo q corresponding to τ under the natural isomorphism ∆ ' (Z/qZ)

×

. Note that p(−τ ) = −p(τ ).

Finally, we let σ be the topological generator of Γ = lim ←− Gal(K

n

/K

0

) which maps ζ

pn

to ζ

p1+pn

for each n ≥ 1 and ζ

q

to ζ

q

. Restrictions of σ to various subfields k

n

, F

n

, F

n+

and K

n

of K

= S

n≥0

K

n

are also denoted by σ.

Now we state the main theorems of this paper:

Theorem 2. If p divides Q

χ∈∆

b

+,χ6=1

B

1,χω−1

, then p divides the class number of F

n+

for all n ≥ 1, where ω is the Teichm¨uller character on Gal(k

0

/Q) ' (Z/pZ)

×

.

Theorem 3. If p does not divide Q

χ∈∆

b

+,χ6=1

B

1,χω−1

, then the prime ideals of F

n+

above p are of order prime to p in the ideal class group of F

n+

. It is well known, by the class number formula, that the relative class number h

K

0

of the field K

0

is given by the formula h

K

0

= Qw Q

%

12

B

1,%

 , where the product is taken over all odd characters of Gal(K

0

/Q). Hence Q

χ∈∆

b

+,χ6=1

B

1,χω−1

contributes to h

K

0

. Thus from Theorems 2 and 3 we obtain some information on the p-divisibility of h

F+

n

= h

+F

n

, the plus part of the class number of F

n

, from that of the minus part h

K

0

.

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In Section 3, we prove a lemma to finish the proof of Theorem 2. This lemma treats certain relations among cyclotomic units. For a deeper analysis of the relations of cyclotomic units, we refer to [2]. We will apply the results of [2] to our special situation. A similar, but slightly different computation was performed in [7].

1. Factorization of a certain principal ideal. In this section, we start out with an explicit element ξ

n

in F

n+

whose norm to F

0+

equals 1. So, by the Hilbert Theorem 90, ξ

n

is of the form ξ

n

= α

σ−1n

for some α

n

∈ F

n+

. The aim of this section is to factorize the principal ideal (α

n

) into a product of prime ideals of F

n+

. This factorization is crucial in the proofs of Theorems 2 and 3 of the following section.

Let

ξ

n

= Y

w∈R

pwn+1

− ζ

q

)(ζ

pwn+1

− ζ

q−1

).

Then ξ

n

is an element of F

n+

since ξ

n

= N

K

n/Fn+

pn+1

− ζ

q

). Since each ζ

pwn+1

− ζ

q±1

is a cyclotomic unit in K

n

, so is ξ

n

. Thus ξ

n

can be thought of as a cyclotomic unit in F

n+

in the sense of W. Sinnott (see [9]). One can easily check that N

F+

n/F0+

n

) = 1. Indeed, N

F+

n/F0+

n

) = N

K

n/F0+

pn+1

− ζ

q

) = N

K

0/F0+

(N

Kn/K0

pn+1

− ζ

q

))

= N

K

0/F0+

p

− ζ

q

) = N

F

0/F0+

 Y

1≤i≤p−1

ζ

pi

− ζ

q



= N

F

0/F0+

 1 − ζ

qp

1 − ζ

q



= 1.

The last equality holds since p ≡ 1 mod q. Hence ξ

n

= α

σ−1n

for some α

n

∈ F

n+

by the Hilbert Theorem 90.

Lemma 1. ξ

n

= α

σ−1n

for some p-unit α

n

∈ F

n+

. P r o o f. Let K

= S

n≥0

Q(ζ

pn+1q

) be the basic Z

p

-extension of K

0

. Let E

0

(C

) be the group of p-units (cyclotomic units) of K

. In [4], Iwa- sawa proves that the cohomology group H

1

(Γ, E

0

) is a finite group, where Γ = Gal(K

/K

0

). But H

1

(Γ, C

) ' (Q

p

/Z

p

)

l

, where l =

12

ϕ(q) (see [5]).

Thus the induced map H

1

(Γ, C

) → H

1

(Γ, E

0

) from the natural inclusion C

→ E

0

must be a zero map. Since the inflation maps on H

1

are injec- tive, H

1

(G

n

, C

n

) → H

1

(G

n

, E

n0

) is a zero map, where G

n

= Gal(K

n

/K

0

) and C

n

(E

n0

) is the group of cyclotomic units (p-units) in K

n

. Then by the cyclicity of the group G

n

, H

−1

(G

n

, C

n

) → H

−1

(G

n

, E

n0

) is also a zero map, which means that a cyclotomic unit in K

n

whose norm to K

0

equals 1 is of the form β

σ−1

for some p-unit β ∈ K

n

. In particular, since

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N

Kn/K0

( Q

w∈R

pwn+1

− ζ

q

)) = 1, we have Y

w∈R

pwn+1

− ζ

q

) = β

σ−1

for some p-unit β ∈ K

n

. Thus

Y

w∈R

pwn+1

− ζ

q

)

pn−1

= N

Kn/Fn

 Y

w∈R

pwn+1

− ζ

q

)



(pn−1)/(p−1)

= N

Kn/Fn

(pn−1)/(p−1)

)

σ−1

. Note that Q

w∈R

pwn+1

− ζ

q

)

pn

= u

σ−1

for some cyclotomic unit u ∈ F

n

since p

n

H

−1

(G

n

, C

Fn

) = 0. Here C

Fn

is the group of cyclotomic units of F

n

. Therefore

Y

w∈R

pwn+1

− ζ

q

) = (uN

Kn/Fn

−(pn−1)/(p−1)

))

σ−1

. Put α

n

= N

F

n/Fn+

(uN

Kn/Fn

−(pn−1)/(p−1)

)). Then ξ

n

= α

σ−1n

and α

n

is a p-unit in F

n+

. This proves the lemma.

Fix a prime ideal ℘

0

of F

0+

above p and let ℘

n

be the prime ideal of F

n+

above ℘

0

. Then {℘

τ0i

| τ

i

∈ ∆

+

} and {℘

τni

| τ

i

∈ ∆

+

} are the sets of all prime ideals of F

0+

and F

n+

above p. For each ℘

τni

, there are two prime ideals in F

n

above ℘

τni

. By abuse of notation, we write them as ℘

τni

and ℘

−τn i

. This will not cause any confusion. Since primes of F

n

above p totally ramify in K

n

, above each prime ideal ℘

±τn i

there is a unique prime ideal e

±τn i

in K

n

. For each τ ∈ ∆ = {±τ

1

, . . . , ±τ

l

}, let F

n,℘τn

be the completion of F

n

at ℘

τn

, and let ϕ

τ

: F

n

→ F

n,℘τn

be the natural embedding. Put s

τ

= ϕ

τ

q

), which is a qth root of 1 in Z

p

. For brevity, we write s for s

id

= ϕ

id

q

). Then s

p(τ )τ

= ϕ

τ

q

)

p(τ )

= ϕ

τ

qp(τ )

) = ϕ

τ

qτ

). Since the completion of F

nτ

at ℘

τn

is the same as the completion of F

n

at ℘

n

, we have s

p(τ )τ

= ϕ

τ

qτ

) = ϕ

id

q

) = s.

Therefore s

τ

= s

p(τ−1)

and s

p(ττ 0)

= s

p(τ−1)p(τ0)

= s

p(τ−1τ0)

for any τ, τ

0

∈ ∆.

Theorem 1. Let ξ

n

be as before and write ξ

n

= α

σ−1n

for some p-unit α

n

∈ F

n+

as in Lemma 1. If (α

n

) = ℘

Σn1≤i≤laiτi

is the factorization in F

n+

, then a

i

≡ 2g(s

p(τi−1)

) mod p, where g(t) is the polynomial defined in the introduction.

R e m a r k. If ξ

n

= α

σ−1n

= (α

0n

)

σ−1

for another p-unit α

0n

, then α

n

= α

n0

α

0

for some p-unit α

0

in F

0+

. Since the primes of F

0+

above p totally ramify in F

n+

, their ramification indices are p

n

. Hence a

i

’s are uniquely determined mod p

n

, so mod p.

One can show that g(s) = g(s

−1

) for any (p − 1)th root s in Z

p

(see [6]).

(5)

Thus

g(s

p(−τi−1)

) = g(s

−p(τi−1)

) = g(s

p(τi−1)

), and g(s

p(τi−1)

) is well defined for each τ

i

∈ ∆

+

.

P r o o f o f T h e o r e m 1. To compute a

i

, we read the prime factoriza- tion (α

n

) = ℘

Σan iτi

in the field K

n, ˜τi

n

(or in K

n, ˜−τi

n

). Since π

n

= ζ

pn+1

− 1 generates the prime ideal of K

n, ˜τi

n

, we have (α

n

) = ℘

Σan iτi

= (π

n

)

ai

. Hence α

n

= π

ani

η for some unit η in K

n, ˜τn

. Note that η

σ−1

≡ 1 mod (π

np

). We claim that π

nσ−1

≡ 1 + π

p−1n

mod (π

pn

). First, notice that for each 1 ≤ k ≤ p − 1, ζ

p1+kpn+1

− 1 = ζ

pn+1

ζ

pkn

− 1 ≡ ζ

pn+1

− 1 mod (ζ

pn

− 1). Thus

Y

1≤k≤p−1

p1+kpn+1

− 1) ≡ Y

1≤k≤p−1

pn+1

− 1)

= (ζ

pn+1

− 1)

p−1

mod (ζ

pn

− 1).

Therefore

π

nσ−1

= ζ

p1+pn+1

− 1

ζ

pn+1

− 1 = ζ

pn

+ ζ

pn

− 1 ζ

pn+1

− 1

= ζ

pn

+ Y

1≤k≤p−1

p1+kpn+1

− 1)

≡ 1 + (ζ

pn+1

− 1)

p−1

mod (ζ

pn

− 1) as claimed. Hence

nai

η)

σ−1

≡ (1 + π

np−1

)

ai

≡ 1 + a

i

π

np−1

mod (π

np

).

On the other hand, by putting x = ζ

pn+1

−1 and t = s

τ

in Q

w∈R

((1 + x)

w

− t) = (1 − t)

p−1

+ g

1

(t)x

p−1

+ (higher terms), we obtain Y

w∈R

pwn+1

− s

τ

) ≡ 1 + g(s

τ

np−1

mod (π

np

).

Hence if we view ξ

n

= Q

w∈R

pwn+1

− ζ

q

)(ζ

pwn+1

− ζ

q−1

) as an element of K

n, ˜℘τn

, we have

ξ

n

= Y

w∈R

pwn+1

− s

τ

)(ζ

pwn+1

− s

−1τ

)

≡ (1 + g(s

τ

np−1

)(1 + g(s

−1τ

p−1n

)

≡ 1 + (g(s

τ

) + g(s

−1τ

))π

np−1

mod (π

np

).

Thus 1 + a

i

π

p−1n

≡ α

σ−1n

= ξ

n

≡ 1 + (g(s

τi

) + g(s

−1τi

))π

np−1

mod (π

np

). There- fore a

i

≡ g(s

τi

) + g(s

−1τi

) ≡ 2g(s

τi

) ≡ 2g(s

p(τi−1)

) mod (π

n

), hence mod p.

2. Main theorems. Recall that l = [F

0+

: Q] and that we arranged

elements of ∆

+

as ∆

+

= {τ

1

, τ

2

, . . . , τ

l−1

, τ

l

= id}. Let A

0

= (a

ij

) be the

(6)

l × l matrix with entries in F

p

such that a

ij

≡ g(s

p(τi−1τj)

) mod p. Note that each row and column of A

0

has a fixed sum g = P

1≤i≤l

g(s

p(τi)

). Let A = (b

ij

) be the l × l matrix with entries in F

p

such that

b

ij

=

 a

ij

if j ≤ l − 1, 1 if j = l.

Below, we write (a

1

, . . . , a

n

)

t

for the column vector with entries a

1

, . . . , a

n

. Lemma 2. Suppose det A ≡ 0 mod p. Then there exists b = (b

1

, . . . , b

l

)

t

in F

lp

such that

(i) A

0

b ≡ 0 = (0, . . . , 0)

t

mod p,

(ii) b is not a constant multiple of 1 = (1, . . . , 1)

t

. P r o o f. We examine the following two cases separately.

C a s e 1: g 6≡ 0 mod p. By adding each column of A

0

to the last one, we have

det A

0

= det

 a

ij j≤l−1

g .. . g

 = g det A ≡ 0 mod p.

Hence A

0

b = 0 has a nontrivial solution. This solution cannot be a multiple of 1, for otherwise, the row sum g would be 0.

C a s e 2: g ≡ 0 mod p. Since each row sum is 0, 1 is obviously a solution of A

0

b = 0. To prove the existence of a solution which is not a multiple of 1, it is enough to check that the F

p

-rank of A

0

is less than or equal to l − 2.

Let B be the (l − 1) × (l − 1) matrix consisting of the first (l − 1) × (l − 1) entries in A

0

. By performing elementary row and column operations, we see that

rank A

0

= rank

 

  B

0 .. . 0 0 . . . 0

0

 

  = rank B.

By adding each row of A to the last one, we have

det A = det

 

  B

1 .. . 1 0 . . . 0

l

 

  = l det B.

Since l 6≡ 0 mod p, det B ≡ 0 mod p. Therefore rank A

0

= rank B ≤ l − 2.

Let ω be the Teichm¨ uller character on Gal(k

0

/Q) ' (Z/pZ)

×

and χ be a

character on ∆

+

. For the proofs of the main theorems, we need the following

(7)

theorem which interprets det A in terms of the generalized Bernoulli numbers B

1,χω−1

.

Theorem. det A ≡ 0 mod p if and only if Q

χ∈∆

b

+,χ6=1

B

1,χω−1

0 mod p.

P r o o f. See Theorem 3 of [7].

Now we restate and prove the main theorems.

Theorem 2. If p | Q

χ∈∆

b

+,χ6=1

B

1,χω−1

, then p | h

F+

n

for all n ≥ 1.

P r o o f. Since h

F+ n

| h

F+

m

for all n ≤ m by the class field theory, it is enough to show that p | h

F+

1

. By the above theorem, we can assume that det A ≡ 0 mod p. Then by Lemma 2, there exists a vector b = (b

1

, . . . , b

l

)

t

in F

lp

satisfying those two conditions (i), (ii) in the lemma. Suppose p - h

F+

1

. Then, by the class field theory, we have

(iii) p - h

F+ 0

.

Moreover, the Sylow p-subgroup of E/C must be trivial (see [9]), where E(C) is the group of units (cyclotomic units) in F

1+

. Thus, the cohomology groups H

i

(G

1

, E/C) are trivial for all i ∈ Z. Hence by considering the long exact sequence of cohomology groups coming from the short exact sequence 0 → C → E → E/C → 0, we have

(iv) The homomorphism H

−1

(G

1

, C) → H

−1

(G

1

, E) induced by the inclusion C → E is an isomorphism.

Let δ = ξ

Σli=1biτi

= α

(Σbiτi)(σ−1)

with b = (b

1

, . . . , b

l

)

t

as before and ξ = ξ

1

, α = α

1

as in Theorem 1. Then the principal ideal (α

Σbiτi

) factorizes as

Σbiτi

) = ℘

(2Σlj=1g(s

p(τ −1 j )

j)(Σli=1biτi)

1

.

In this expression,

 X

l

j=1

g(s

p(τj−1)

j

 X

l

i=1

b

i

τ

i



= X

1≤i,j≤l

g(s

p(τj−1)

)b

i

τ

j

τ

i

= X

l k=1

 X

τjτi,jik

g(s

p(τj−1)

)b

i

 τ

k

= X

l k=1

 X

l

i=1

g(s

p(τk−1τi)

)b

i

 τ

k

.

Since A

0

b ≡ 0, P

l

i=1

g(s

p(τk−1τi)

)b

i

≡ 0 mod p for each k = 1, . . . , l. Hence

Σbiτi

) = ℘

Σ1li=1diτi

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for some d

i

satisfying d

i

≡ 0 mod p. Since ℘

p1

= ℘

0

, we get

Σbiτi

) = I

0

for some ideal I

0

of F

0+

. But the subgroup of the ideal class group of F

0+

consisting of ideal classes that become principal in F

1+

is a p-group. Thus nonprincipal ideals of F

0+

cannot capitulate in F

1+

by (iii). Therefore

Σbiτi

) = I

0

= (α

0

)

for some α

0

∈ F

1+

, and thus α

Σbiτi

= α

0

η

0

for some unit η

0

in F

1+

. Then we have

δ = ξ

Σbiτi

= α

(Σbiτi)(σ−1)

= (η

0

)

σ−1

.

Since the induced homomorphism H

−1

(G

1

, C) → H

−1

(G

1

, E) is an isomor- phism by (iv), (η

0

)

σ−1

= η

σ−1

for some cyclotomic unit in F

1+

. However, this cannot happen by (ii) and the following lemma which will be proved in the next section.

Lemma 3. Let ξ = Q

w∈R

pw2

− ζ

q

)(ζ

pw2

− ζ

q−1

) as before. If ξ

Σli=1ciτi

= η

σ−1

for some cyclotomic unit η ∈ F

1+

, then c

1

≡ . . . ≡ c

l

mod p.

Theorem 3. If p - Q

χ∈∆

b

+,χ6=1

B

1,χω−1

, then the prime ideals of F

n+

above p are of order prime to p in the ideal class group of F

n+

for all n ≥ 0.

P r o o f. It is enough to show that the ideal class [℘

n

] is of order prime to p. As in Lemma 2, we examine two cases separately.

C a s e 1: g 6≡ 0 mod p. Since det A

0

= g det A, det A

0

6≡ 0 mod p. Thus there exists x = (x

1

, . . . , x

l

)

t

such that A

0

x = (0, . . . , 0, 1)

t

. Let ξ

n

and α

n

be as in Theorem 1. Then (α

Σnli=1xiτi

) = ℘

Σnli=1diτi

for some d

i

satisfying d

l

≡ 1 mod p and d

1

≡ . . . ≡ d

l−1

≡ 0 mod p. Since ℘

pn

= ℘

n−1

, we get (∗)

Σxn iτi

) = ℘

n

I

n−1

for some ideal I

n−1

of F

n−1+

whose prime factors lie above p. Let mp

k

be the order of the ideal class [℘

n

] with (m, p) = 1. If k 6= 0, then I

n−1mpk−1

is a principal ideal in F

n+

. Therefore by raising both sides of (∗) to the power of dp

k−1

we get a contradiction. Hence k = 0.

C a s e 2: g ≡ 0 mod p. Since det A 6≡ 0 mod p, F

p

-rank A

0

= l − 1.

Thus columns of A

0

except the last one are linearly independent over F

p

and are contained in the subspace of F

lp

consisting of {y = (y

1

, . . . , y

l

)

t

F

lp

| y

1

+ . . . + y

l

= 0}, which is of dimension l − 1. Therefore if we view

A

0

as a linear map from F

lp

to F

lp

, the image of A

0

is precisely the subspace

described above. For each i, 1 ≤ i ≤ l − 1, choose b

i

in F

lp

such that

A

0

b

i

= (0, . . . , 1, . . . , 0, −1)

t

, with 1 at the ith place, −1 at the last place

(9)

and 0 elsewhere. Then as in the first case, we get (∗∗)

τni−1

I

n−1

= (β

n

)

for some β

n

∈ F

n+

. Note that ℘

Σnli=1τi

is the prime ideal of Q

n

, hence is principal. Thus by multiplying (∗∗) for 1 ≤ i ≤ l we see that ℘

ln

I

n−10

is a principal ideal for some ideal I

n−10

whose prime factors lies above p. Since p - l, we can check that [℘

n

] is of order prime to p as in Case 1.

Let A

n

be the Sylow p-subgroup of the ideal class group of F

n+

and let A

= lim ←− A

n

, where the limit is taken under the norm maps. It is well known that A

' Z

λp

⊕M for some finite group M which measures the capitulation.

It is conjectured that the Iwasawa λ-invariant for F

+

/F

0+

equals 0 and R. Greenberg gave several equivalent statements to this (see [3]). By using one of the equivalent statements (Theorem 2 of [3]), we have the following corollary.

Corollary. Suppose λ = 0. Then p - Q

χ∈∆

b

+,χ6=1

B

1,χω−1

if and only if A

= {0}.

P r o o f. Theorem 2 takes care of the if part. For the converse, we need the assumption λ = 0. Since λ = 0, #A

n

is bounded by #M as n → ∞.

Equivalently, every ideal class in A

Gnn

contains an ideal whose prime factors lie above p by Theorem 2 of [3]. Thus if p - Q

χ∈∆

b

+,χ6=1

B

1,χω−1

, then A

Gnn

= {0} by Theorem 3. Since A

n

and G

n

are p-groups, A

n

= {0} for all n.

Therefore A

= {0}.

3. Proof of Lemma 3. In this section we prove Lemma 3 stated in the previous section. The proof is based on the work of V. Ennola ([2]) and is similar to that of Theorem 1 in [7]. In particular we need the following theorem:

Theorem (V. Ennola). Let δ = Q

1≤a<n

(1 − ζ

na

)

xa

, x

a

∈ Z, be a cyclo- tomic unit in Q(ζ

n

). For an even character χ 6= 1 of conductor f belonging to Q(ζ

n

), define Y (χ, δ) by

Y (χ, δ) = X

f |d|nd

1

ϕ(d) T (χ, d, δ) Y

p|d

(1 − χ(p)), where

T (χ, d, δ) =

d−1

X

(a,d)=1a=1

χ(a)x

na/d

.

If δ is a root of 1, then Y (χ, δ) = 0 for all even characters χ 6= 1 belonging

to Q(ζ

n

).

(10)

We also need the following properties of Y which are easy to check. Let χ 6= 1 be an even character belonging to Q(ζ

n

). Then

(i) Y (χ, δ

1

δ

2

) = Y (χ, δ

1

) + Y (χ, δ

2

).

(ii) If (root of 1)×δ

1

= (root of 1)×δ

2

, then Y (χ, δ

1

) = Y (χ, δ

2

).

(iii) For any σ ∈ Gal(Q(ζ

n

)/Q), Y (χ, δ

σ

) = χ(σ)Y (χ, δ).

(iv) Y (χ, δ

σ−1

) = (χ(σ) − 1)Y (χ, δ).

Now we sketch the proof of Lemma 3 briefly.

S k e t c h o f p r o o f. Suppose ξ

Σli=1ciτi

= η

σ−1

, with ξ, η as in the lemma. By (ii),

Y (%, ξ

Σciτi

) = Y (%, η

σ−1

)

for any even character % 6= 1 in Gal(K

1

/Q)

. By (i), (iv), we obtain (∗)

X

l i=1

c

i

%(τ

i

)Y (%, ξ) = (%(σ) − 1)Y (%, η).

Fix a nontrivial character ψ of Gal(Q

1

/Q). For a nontrivial character χ ∈

b

+

, put % = χψ in (∗). After a similar computation to that of Theorem 1 of [7], we have

(p − 1) X

l i=1

c

i

χ(τ

i

) = (ψ(σ) − 1)α(χ)

for some algebraic integer α(χ). By letting χ run through all the nontriv- ial even characters of b

+

, we have the following system of linear equa- tions:

(p − 1)T (c

1

, . . . , c

l

)

t

= (ψ(σ) − 1)(. . . , α(χ), . . .)

t

,

where T is the (l − 1) × l matrix with rows of the form (χ(τ

1

), . . . , χ(τ

l

)).

Let L = Q(ζ

p

, α(χ)’s, χ(τ

i

)’s), O

L

be the ring of integers of L, and P be a prime ideal of O

L

above p. Then we have

T (c

1

, . . . , c

l

)

t

≡ (0, . . . , 0)

t

mod P, since ψ(σ) − 1 ≡ ζ

p

− 1 ≡ 0 mod P.

Let T be the matrix obtained by reducing the entries of T mod P. By

using Lemma 1.2 of [7], one can check that the O

L

/P-rank of T is l−1. Hence

{x = (x

1

, . . . , x

l

)

t

∈ O

L

/P | T x = (0, . . . , 0)

t

} is one-dimensional. But

(1, . . . , 1)

t

obviously satisfies T (1, . . . , 1)

t

= (0, . . . , 0)

t

. Thus (c

1

, . . . , c

l

)

t

α(1, . . . , 1)

t

mod P for some α ∈ O

L

. Therefore c

1

≡ . . . ≡ c

l

mod P, hence

mod p.

(11)

References

[1] J. C o a t e s and A. W i l e s, On p-adic L-functions and elliptic units, J. Austral.

Math. Soc. 26 (1978), 1–25.

[2] V. E n n o l a, On relations between cyclotomic units, J. Number Theory 4 (1972), 236–247.

[3] R. G r e e n b e r g, On the Iwasawa invariants of totally real number fields, Amer. J.

Math. 98 (1976), 263–284.

[4] K. I w a s a w a, On cohomology groups of units for Z

p

-extensions, ibid. 105 (1983), 189–200.

[5] J. M. K i m, Cohomology groups of cyclotomic units, J. Algebra 152 (1992), 514–519.

[6] —, Coates–Wiles series and Mirimanoff’s polynomial, J. Number Theory, to ap- pear.

[7] —, Units and cyclotomic units in Z

p

-extensions, Nagoya Math. J. (1995), to appear.

[8] P. R i b e n b o i m, 13 Lectures on Fermat’s Last Theorem, Springer, New York, 1972.

[9] W. S i n n o t t, On the Stickelberger ideal and the circular units of a cyclotomic field, Ann. of Math. 108 (1978), 107–134.

[10] L. W a s h i n g t o n, Introduction to Cyclotomic Fields, Graduate Texts in Math. 83, Springer, New York, 1980.

DEPARTMENT OF MATHEMATICS INHA UNIVERSITY, INCHON, KOREA E-mail: JMKIM@MUNHAK.INHA.AC.KR

Received on 17.8.1994

and in revised form on 2.11.1994 (2656)

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